字符串修改

问题描述

大家好。我无法理解为什么以下程序无法正常工作。请解释。


int main（）

{

char * t =" hello"，* r;

r = t;

* r =''H'';

printf（"％s"，t）;

返回0;

}

导致分段错误.Bye 。


问候，

Jerico

Hi all.I could not understand why the following program didn''t work.Please explain.

int main()
{
char *t="hello",*r;
r=t;
*r=''H'';
printf("%s",t);
return 0;
}
It results in segmentation fault.Bye.

Regards,
Jerico

推荐答案

问题在于分配给你好"在程序数据段中，因此受到保护，任何修改它的尝试都会导致分段错误。您可以使用strdup函数来解决这个问题。这将在堆中创建一个可以自由修改的副本。

The problem is that the memory allocated for "hello" is in the programs data segment and is therefore protected and any attempt to modify it will result in a segmentation fault.. You can get around this by using the strdup function. This will create a duplicate in the heap that you can freely modify.

展开 | 选择 | Wrap | 行号

char t [] =" hello"，* r;

char t[]="hello",*r;

r = t;

* r =''H'';

printf（"％s"，t）;

返回0;

}

r=t;
*r=''H'';
printf("%s",t);
return 0;
}

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